CAIE M1 2011 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2011
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeResultant of three coplanar forces
DifficultyModerate -0.3 This is a standard M1 mechanics question requiring resolution of forces into components and finding the resultant. While it involves multiple forces and angles, the method is routine: resolve each force into horizontal/vertical (or along AB/perpendicular) components, sum them, then use Pythagoras and trigonometry. The angles are straightforward (30°, 60°, 90°), making calculations cleaner than average. Slightly easier than a typical A-level question due to its procedural nature and nice angles.
Spec3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors

3 \includegraphics[max width=\textwidth, alt={}, center]{28562a1b-ec9a-40d2-bbb3-729770688971-2_476_714_744_719} Three coplanar forces of magnitudes \(15 \mathrm {~N} , 12 \mathrm {~N}\) and 12 N act at a point \(A\) in directions as shown in the diagram.
  1. Find the component of the resultant of the three forces
    1. in the direction of \(A B\),
    2. perpendicular to \(A B\).
    3. Hence find the magnitude and direction of the resultant of the three forces.

AnswerMarks Guidance
(i) (a) \([2 \times 12\cos 40° - 15\cos 50°]\)M1 For resolving in direction \(\overrightarrow{AB}\)
Component is 8.74 NA1
(b) Component is 11.5 NB1 3
(ii) Magnitude is 14.4 N or direction is \(52.7°\) (or \(0.920°\)) anticlockwise from \(\mathbf{i}\) dir'nM1 For using \(R^2 = X^2 + Y^2\) or \(\tan\theta = Y/X\)
A1
Direction is \(52.7°\) (or \(0.920°\)) anticlockwise from \(\mathbf{i}\) dir'n or magnitude is 14.4 NB1 3
| **(i) (a)** $[2 \times 12\cos 40° - 15\cos 50°]$ | M1 | For resolving in direction $\overrightarrow{AB}$ |
| Component is 8.74 N | A1 | |
| **(b)** Component is 11.5 N | B1 | 3 |
| **(ii)** Magnitude is 14.4 N or direction is $52.7°$ (or $0.920°$) anticlockwise from $\mathbf{i}$ dir'n | M1 | For using $R^2 = X^2 + Y^2$ or $\tan\theta = Y/X$ |
| | A1 | |
| Direction is $52.7°$ (or $0.920°$) anticlockwise from $\mathbf{i}$ dir'n or magnitude is 14.4 N | B1 | 3 |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{28562a1b-ec9a-40d2-bbb3-729770688971-2_476_714_744_719}

Three coplanar forces of magnitudes $15 \mathrm {~N} , 12 \mathrm {~N}$ and 12 N act at a point $A$ in directions as shown in the diagram.\\
(i) Find the component of the resultant of the three forces
\begin{enumerate}[label=(\alph*)]
\item in the direction of $A B$,
\item perpendicular to $A B$.\\
(ii) Hence find the magnitude and direction of the resultant of the three forces.
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2011 Q3 [6]}}